How to use symbolic toolbox to remove a redundant variable from the equation?

Question

Rui 2017-10-7

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https://ww2.mathworks.cn/matlabcentral/answers/360180-how-to-use-symbolic-toolbox-to-remove-a-redundant-variable-from-the-equation

评论： Rui 2017-10-10

Dear all,

I have two equations y1 = F1(x,a,b) and y2 = F2(x,m,n), where x is an independent variable, while a, b, m and n are constants.

Assuming y1 = y2, is there a way to use symbolic toolbox to remove x from above equations, and get m and n as functions of a and b (i.e. m = G1(a,b), n = G2(a,b)) ?

Thank you!

Rui

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Answer 1

Kian Azami 2017-10-7

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maybe if you put your functions here it will be better. However, logically you have two equations and three unknowns and it is not possible to find two unknown separate than x.

Consider you find the value of x from the first equation which will be equal to something as:

x = x(y1,a,b). if y1 == y2 => x = x(y2,a,b)

Then you put this in the second equation:

y2 = F2(x(y2,a,b),m,n)

Now you only have one equation and you want to find two values which are not possible.

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Rui 2017-10-7

在 MATLAB Online 中打开

Thanks, Kian! Here is the full problem ...

I have

response = Emax_AR *( (C_A*C_R_total*koff_B^2*kon_A + C_A^2*C_R_total*koff_B*kon_A^2 + C_A*C_B*C_R_total*koff_B*kon_A*kon_B + C_A*C_R_total*alpha*koff_A*koff_B*kon_A)/(koff_A*koff_B^2 + C_A^2*koff_B*kon_A^2 + C_A*koff_B^2*kon_A + alpha*koff_A^2*koff_B + C_B*alpha*koff_A^2*kon_B + C_A*C_B^2*kon_A*kon_B^2 + C_A^2*C_B*kon_A^2*kon_B + C_B^2*alpha*koff_A*kon_B^2 + C_A*koff_A*koff_B*kon_A + C_B*koff_A*koff_B*kon_B + C_A*C_B*koff_A*kon_A*kon_B + 2*C_A*C_B*koff_B*kon_A*kon_B + C_A*alpha*koff_A*koff_B*kon_A + C_B*alpha*koff_A*koff_B*kon_B + C_A*C_B*alpha*koff_A*kon_A*kon_B)) + ...
Emax_ARB * ((C_A*C_B^2*C_R_total*kon_A*kon_B^2 + C_A^2*C_B*C_R_total*kon_A^2*kon_B + C_A*C_B*C_R_total*koff_A*kon_A*kon_B + C_A*C_B*C_R_total*koff_B*kon_A*kon_B)/(koff_A*koff_B^2 + C_A^2*koff_B*kon_A^2 + C_A*koff_B^2*kon_A + alpha*koff_A^2*koff_B + C_B*alpha*koff_A^2*kon_B + C_A*C_B^2*kon_A*kon_B^2 + C_A^2*C_B*kon_A^2*kon_B + C_B^2*alpha*koff_A*kon_B^2 + C_A*koff_A*koff_B*kon_A + C_B*koff_A*koff_B*kon_B + C_A*C_B*koff_A*kon_A*kon_B + 2*C_A*C_B*koff_B*kon_A*kon_B + C_A*alpha*koff_A*koff_B*kon_A + C_B*alpha*koff_A*koff_B*kon_B + C_A*C_B*alpha*koff_A*kon_A*kon_B))

C_B is the independent variable, I know this equation follows a sigmoid curve which has a general format as

response = Emax*(1 - C_B / (C_B + IC50) )

I would like to define IC50 and Emax using constants in the first equation.

Kian Azami 2017-10-7

But why you don't put the whole code. I mean without the syms variable that you defined for defining the "response" the lines you put here is unreadable.

Rui 2017-10-10

Hi Kian, sorry for confusion. I have solved the issue. For sigmoid equation, (1) IC50 is defined as the C_B concentration where the response is half of the max response. (2) The max response will be reached when C_B is zero. As such, I substituted all C_B in the first equation with 0 to get the max response. Then I made max response / 2 == response, and solve for C_B. At the end, I got a C_B value where the response is half of max response, this C_B value is my IC50.

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